# Lambda Calculus - are these two expressions equivalent? [closed]

(λa.(λb.λc.b)) and (λa.λb.λc.b)

I was wondering if someone could explain, using mostly English, what that lambda-calculus expression is supposed to mean, and whether there is any difference between the two given above.

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## closed as off topic by Artem Kaznatcheev♦, Jukka Suomela, KavehMar 13 '12 at 5:06

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The above question would be a better fit at Mathematics – Artem Kaznatcheev Jan 27 '12 at 11:21
Welcome to cstheory, a Q&A site for research-level questions in theoretical computer science (TCS). Your question does not appear to be a research-level question in TCS. Please see the FAQ for more information on what is meant by this and suggestions for sites that might welcome your question. Finally, if your question is closed for being out of scope, and you believe you can edit the question to make it a research-level question, please feel free to do so. Closing is not permanent and questions can be reopened, check the FAQ for more information. – Kaveh Mar 13 '12 at 5:05

It's the same. In $λa.(λb.(λc.b))$ brackets are not mandatory because there is no other way to parse $λa.λb.λc.b$ with λ-calculus syntax:
$$t ::= x \mid λx.t \mid t\,t$$